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frez [133]
1 year ago
7

star a and star b have measured stellar parallax of 0.58 arc second and 0.73 arc second, respectively. which star is closer?

Physics
1 answer:
goldenfox [79]1 year ago
3 0

Star B is closer than star A, because 1/ 0.58 = 1.72 pc, while 1/.73 = 1.36 pc.

<h3>What is stellar parallax and distance measurement using stellar parallax?</h3>

To calculate the distances to nearby stars, astronomers employ a phenomenon known as parallax. The apparent displacement of an object due to a change in the viewer's point of view is known as parallax. The distances between close stars can be calculated using this phenomenon. A nearer star will seem to move against the farther-off background stars as the Earth revolves around the Sun. By measuring a star's position once, then again six months later, astronomers can establish the apparent shift in location of that star. The apparent motion of the star is referred to as stellar parallax.

A straightforward correlation exists between the distance of a star and parallax angle:

d=\frac{1}{p}

The distance d and parallax angle p are both measured in parsecs and arcseconds, respectively.

Stellar parallax of star A= 0.58 arcseconds

Stellar parallax of star B= 0.73 arcseconds

So, on applying the formula,

The distance of the star A= 1.72pc

And the distance of the star B= 1.36pc

Therefore, star B is closer than star A.

To know more about stellar parallax, visit:

brainly.com/question/29352944

#SPJ4

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A sinusoidal voltage is given by the expression ????(????)=20cos(5π×103 ????+60°) V. Determine its (a) frequency in hertz, (b) p
MA_775_DIABLO [31]

<em>There are some placeholders in the expression, but they can be safely assumed</em>

Answer:

(a) f=1617.9\ Hz

(b) T=0.618\ ms

(c) A=20 \ Volts

(d) \varphi=60^o

Explanation:

<u>Sinusoidal Waves </u>

An oscillating wave can be expressed as a sinusoidal function as follows

V(t)&=A\cdot \sin(2\pi ft+\varphi )

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A=Amplitude

f=frequency

\varphi=Phase\  angle

The voltage of the question is the sinusoid expression  

V(t)=20cos(5\pi\times 103t+60^o)

(a) By comparing with the general formula we have

f=5\pi\times 103=1617.9\ Hz

\boxed{f=1617.9\ Hz}

(b) The period is the reciprocal of the frequency:

\displaystyle T=\frac{1}{f}

\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec

Converting to milliseconds

\boxed{T=0.618\ ms}

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4 0
3 years ago
The conducting path between the right hand and the left hand can be modeled as a 9.0-cm-diameter, 140-cm-long cylinder. The aver
Crank

Answer:

The potential difference is 121.069 V

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As per the question:

Diameter of the cylinder, d = 9.0 cm = 0.09 m

Length of the cylinder, l = 40 cm = 1.4 m

Average Resistivity, \rho = 5.5\ \Omega-m

Current, I = 100 mA = 0.1 A

Now,

To calculate the potential difference between the hands:

Cross- sectional Area of the Cylinder, A = \pi (\frac{d}{2})^{2} = 6.36\times 10^{- 3}\ m^{2}

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V = 0.1\times 1210.69 = 121.069\ V

4 0
4 years ago
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